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A242879
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Least positive integer k < n such that k*p == 1 (mod prime(k)) for some prime p < prime(k) and (n-k)*q == 1 (mod prime(n-k)) for some prime q < prime(n-k), or 0 if such a number k does not exist.
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1
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0, 0, 0, 2, 2, 2, 3, 2, 2, 3, 4, 2, 2, 3, 2, 3, 4, 7, 2, 2, 3, 4, 2, 3, 4, 13, 6, 7, 11, 13, 10, 11, 2, 3, 4, 18, 6, 7, 2, 3, 4, 2, 2, 3, 4, 6, 6, 2, 3, 2, 2, 3, 4, 2, 2, 3, 4, 6, 6, 2, 3, 2, 3, 4, 7, 2, 3, 2, 3, 4, 7, 2, 2, 2, 2, 3, 2, 3, 4, 7
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OFFSET
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1,4
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COMMENTS
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According to the conjecture in A242753, a(n) should be positive for all n > 3.
We have verified that a(n) > 0 for all n = 4, ..., 10^8.
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LINKS
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EXAMPLE
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a(4) = 2 since 4 = 2 + 2 and 2*2 == 1 (mod prime(2)=3).
a(7) = 3 since 7 = 3 + 4, 3*2 == 1 (mod prime(3)=5) with 2 prime, and also 4*2 == 1 (mod prime(4)=7) with 2 prime, but 5*9 == 1 (mod prime(5)=11) with 9 not prime.
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MATHEMATICA
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p[n_]:=PrimeQ[PowerMod[n, -1, Prime[n]]]
Do[Do[If[p[k]&&p[n-k], Print[n, " ", k]; Goto[aa]]; Continue, {k, 1, n/2}]; Print[n, " ", 0]; Label[aa]; Continue, {n, 1, 80}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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